1. ## Differential equations substitution

Solve:

$\displaystyle 2y'y''=1+(y')^2$ by letting $\displaystyle y'=v$

Does $\displaystyle y''=v'$

So does $\displaystyle 2vv'=1+v^2$

2. Right. So what can you do now?

3. What is $\displaystyle v'$ with respect to?

4. The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.

5. Originally Posted by Ackbeet
The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.
So, $\displaystyle \dfrac{2v}{1+v^2} \ dv=dx$

$\displaystyle \displaystyle\int{\dfrac{2v}{1+v^2} \ dv}=\int{dx}$

$\displaystyle \ln|1+v^2|=x+C$

$\displaystyle 1+v^2=Ce^x$

$\displaystyle v^2=Ce^x-1$

6. Not sure I would use the division symbol there. On the LHS, you have $\displaystyle dv/dx$, originally. When you multiply both sides of

$\displaystyle \displaystyle{\frac{2v}{1+v^{2}}\,\frac{dv}{dx}=1}$ by $\displaystyle dx$, what are you going to get?

[EDIT]: acevipa edited post # 5. It's correct now.

7. Didn't I do it correct?

8. Oh, I see. You edited your post. That looks good to me.

9. Originally Posted by Ackbeet
Oh, I see. You edited your post. That looks good to me.
Therefore, $\displaystyle (y')^2=Ce^x-1$

What do you do from here? Don't we have to find y?